Does there exist a rational function modulo $2^n$? This is a function of the form $$f(a,b)\pmod{2^n},$$ such that $$\forall\;a,b\in \{0,1,\ldots,2^n-1\},\quad a\oplus b=f(a,b)\pmod{2^n}.$$

It's trivial that when $n=1,2$, we can use $$f(a,b)=a+b-2ab\pmod{2^n}.$$

Does there exist a rational function modulo $2^n$? This is a function of the form $$f(a,b)\pmod{2^n},$$ such that $$\forall\;a,b\in \{0,1,\ldots,2^n-1\},\quad a\oplus b=\frac{g(a,b)}{h(a,b)}\pmod{2^n},$$ where $g$ and $h$ are polynomials.

It's trivial that when $n=1,2$, we can use $$f(a,b)=a+b-2ab\pmod{2^n}.$$

Whether there exists a rational function modulue $2^n$: $f(a,b)(mod\;2^n)$, s.t. $\forall\;a,b\in \{0,1,...,2^n-1\}$, $a\oplus b=f(a,b)(mod\;2^n)$

It is trivial that when $n=1,2$, $f(a,b)=a+b-2ab(mod \;2^n)$ is what we need.

Does there exist a rational function modulo $2^n$? This is a function of the form $$f(a,b)\pmod{2^n},$$ such that $$\forall\;a,b\in \{0,1,\ldots,2^n-1\},\quad a\oplus b=f(a,b)\pmod{2^n}.$$

It's trivial that when $n=1,2$, we can use $$f(a,b)=a+b-2ab\pmod{2^n}.$$